India’s No.1 Educational Platform For UPSC,PSC And All Competitive Exam
• 0
• Donate Now
Probability
1.Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Explanation:

Solution:
Here, S = {1, 2, 3, 4, ...., 19, 20}.
Let E = event of getting a multiple of 3 or 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{9}{20}$

2.A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

Explanation:

Solution:
$\begin{array}{rl}& \text{Total}\phantom{\rule{thinmathspace}{0ex}}\text{number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{balls}\\ & =\left(2+3+2\right)=7.\\ & \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{S}\phantom{\rule{thinmathspace}{0ex}}\text{be}\phantom{\rule{thinmathspace}{0ex}}\text{the}\phantom{\rule{thinmathspace}{0ex}}\text{sample}\phantom{\rule{thinmathspace}{0ex}}\text{space}\text{.}\\ & \text{Then,}\phantom{\rule{thinmathspace}{0ex}}n\left(S\right)=\text{Number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{ways}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{drawing}\phantom{\rule{thinmathspace}{0ex}}\text{2}\phantom{\rule{thinmathspace}{0ex}}\text{balls}\phantom{\rule{thinmathspace}{0ex}}\text{out}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{7}\\ & ={}^{7}{C}_{2}\\ & =\frac{\left(7×6\right)}{\left(2×1\right)}\\ & =21.\\ & \text{Let}\phantom{\rule{thinmathspace}{0ex}}\text{E = Event}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{drawing}\phantom{\rule{thinmathspace}{0ex}}\text{2}\phantom{\rule{thinmathspace}{0ex}}\text{balls,}\phantom{\rule{thinmathspace}{0ex}}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{none}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{which}\phantom{\rule{thinmathspace}{0ex}}\text{is}\phantom{\rule{thinmathspace}{0ex}}\text{blue}\text{.}\\ & \therefore n\left(E\right)=\text{Number}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\text{ways}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{drawing}\phantom{\rule{thinmathspace}{0ex}}\text{2}\phantom{\rule{thinmathspace}{0ex}}\text{balls}\phantom{\rule{thinmathspace}{0ex}}\text{out}\phantom{\rule{thinmathspace}{0ex}}\text{of}\phantom{\rule{thinmathspace}{0ex}}\left(\text{2 + 3}\right)\text{balls}\\ & ={}^{5}{C}_{2}\\ & =\frac{\left(5×4\right)}{\left(2×1\right)}\\ & =10.\\ & \therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{21}.\end{array}$

3.In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Explanation:

Solution:

4.Three unbiased coins are tossed. What is the probability of getting at most two heads?

Explanation:

Solution:
Getting at most Two heads means 0 to 2 but not more than 2.
Here S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}.
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{7}{8}$

5.Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Explanation:

Solution:
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.

Then, E= {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
$\begin{array}{rl}& \therefore n\left(E\right)=27\\ & \therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ & =\frac{27}{36}=\frac{3}{4}\end{array}$

6.In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

Explanation:

Solution:
Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25.
$\begin{array}{rl}& ={}^{25}{C}_{3}\\ & =\frac{\left(25×24×23\right)}{\left(3×2×1\right)}\\ & =2300.\\ & n\left(E\right)=\left({}^{10}{C}_{1}×{}^{15}{C}_{2}\right)\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\left[10×\frac{\left(15×14\right)}{\left(2×1\right)}\right]\\ & \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=1050.\\ & \therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{1050}{2300}=\frac{21}{46}.\end{array}$

7.In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Explanation:

Solution:
$\begin{array}{rl}& P\left(\text{getting}\phantom{\rule{thinmathspace}{0ex}}\text{a}\phantom{\rule{thinmathspace}{0ex}}\text{prize}\right)\\ & =\frac{10}{10+25}\\ & =\frac{10}{35}\\ & =\frac{2}{7}\end{array}$

8.Two dice are tossed. The probability that the total score is a prime number is:

Explanation:

Solution:
Clearly, n(S) = (6 x 6) = 36.
Let E = Event that the sum is a prime number.Then
E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5) }
$\begin{array}{rl}& \therefore n\left(E\right)=15.\\ & \therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ & =\frac{15}{36}\\ & =\frac{5}{12}\end{array}$

9.A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:

Explanation:

Solution:
Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart.
Then, n(E) = 2.
$\begin{array}{rl}& \therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}\\ & =\frac{2}{52}\\ & =\frac{1}{26}\end{array}$

10.A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is: